A cyclist leaves her home with a bearing of 150.

Question

After cycling for 10​ miles, she heads due east for another 10 miles. To get back home from her current​ position, the cyclist must follow a bearing of 300 degrees. How far must she pedal to get​ home?

Answer 1

So, just out of the shoot, a bearing is based on a compass, North is both 0°/360°, clockwise is positive bearings. Next draw a picture (which is hard to do here). So here is the description.

1. from N,S,E,W compass (North = 0, + degrees CW), draw first leg 150°, which will be in the SE direction.
2. Label 10 on that leg length.
3. From that point on the leg, go due East = 90°, again for 10 miles, and label that.
4. From that point back to your starting point (origin), draw a line, and put X for the length. Now you will have a triangle with the following degrees (from top CW inside the triangle):
5. 30°, 30°, 120°; and associated legs, top CW would be X, 10, 10 (I hope that makes sense to you)
6. To solve for X (the hypotenuse of the triangle, as it is across from the largest angle), you would use a trig function such as: sin 30 = 10/X and solve for X
7. X = 20

Answer 2

This is a question with bearings and trigonometry, great fun!

Always best to start with a sketch so we know what the triangle looks like.

Bearing is clockwise from vertical, so 0 degrees = North.

Let's start at H for Home, then first leg to A (side HA), then second leg to B (side AB), and then return home (side BH). We need the length of BH, or distance to home.

Oops! Turns out I can't include my sketch! That makes my labelling harder to follow, sorry. Hopefully, you can figure out what I'm drawing!

The triangle HAB that we're interested in, is bounded by a rectangle: two parallel vertical lines (North), one passing through H. And two parallel horizontal lines (East), one passing through H, and the other the second leg. You can see there are 3 triangles (and HAB can be divided in 2), so there are several different ways of finding the solution, depending on the trigonometry you are most comfortable with.

We know HA = HB = 10. That means HAB is an isosceles triangle, so I think it's easiest to solve by dividing into 2 right angle triangles.

We have a triangle with two known sides. And we need to find the remaining side, the home leg. For triangle problems, we must know 3 things to be able to solve it, so this tells us we should be able to work out at least 1 of the angles (although I will try to find all 3, even though it's not strictly necessary). So let's do that...

Angle N(orth)HA is our initial bearing 150 degrees. This is 90 (to east) plus 60 more for E(east)HA.

Second leg is BA due east.

Third leg back BH at 300 degrees, which means angle N(orth)BH = 60 degrees.

As you can see, two parallel lines (east) with a crossing line, or "transversal", (HA) tells us that angle HAZ (Z is due south of H, on the East line passing through AB) is also 60 degrees ("alternate interior angles are equal", or the "Z").

If angle HAZ is 60 degrees, then angle HAB must be 120 degrees ("angles on a straight line"). This is the main angle we need of our route triangle HAB.

If angle HAB is 120 degrees then the remaining angles must be 60 degrees ("angles in a triangle"). And as we already know HAB is isoceles then we can say angle ABH = angle BHA = 30 degrees.

If you're smarter than me (and wait until you draw the third leg), you can see this leg is 60 degrees before north, which means 30 degrees to make a right angle with your second (due east) leg.

Whichever way you work it out, you have an isosceles triangle with 30/120/30 and two short sides of 10.

Working out the long side depends on whether you know trig formulae for right angled triangles, or any angle triangles.

Splitting the triangle HAB into two equal right angle triangles, we can see that AB is now a hypotenuse with length 10, while the opposite side is unknown, and the adjacent side is unknown, but is half the length BH that we're looking for. Adjacent/hypotenuse is cosine (using standard trig function in right angle triangle).

Hence, adjacent/10 = cosine (30 deg)

=> adjacent = 10 cosine (30 deg) = 8.66

Finally, BH = 2 x 8.66 = 17.3 miles

Another way to get this with right angle maths, is by working out the AZ line (where Z due south of H). This is a right angled triangle (HAZ), so adjacent/hypotenuse = cos (60 deg) => 10 cos (60 deg) = adjacent = 5.

Then add 5 (AZ) to 10 (AB) to get the base (15) of another right angle triangle HZB, and so adjacent/hypotenuse = cos (30 deg) => hypotenuse BH = 15 / cos (30 deg) = 17.3 miles again.

OR... If you like non-right angle maths, you can also do directly with triangle HAB, using the sine rule.

sin a / A = sin b / B = sin c / C

sin b / B = sin (30 deg) / 10 = sin c / C

but sin a / A = sin (120 deg) / BH

Hence, sin (120 deg) / BH = sin (30 deg) / 10

=> 10 sin (120 deg) / sin (30 deg) = BH

=> BH = 17.3 miles

OR... if you prefer, you can use the cosine rule (sq rt of (a² + b² - 2ab cos c)) with triangle HAB.

In this case, BH = √ of (10² + 10²) - 2(10)(10) cos (120 deg) = √ (200 - 200 cos (120 deg)) = 17.3 miles

As a little reality check, a right angle triangle (less than our actual 120 degrees) with two sides of 10, 10 would have a hypotenuse of sq rt (10² + 10²) = √(200) = 14 ish. So our length to home must be longer than this. 120 deg vs 90 deg is about 30% bigger, which means a guesstimate would be 18 ish. Pretty close to our answer of 17.3.